Quotient Object of Subobjects A problem (not homework) from CWM:

For subobjects $u\leq v$ of an object $a$ in an abelian category $\mathsf A$, define a "quotient" object $v/u$ (to agree with the usual notion in $\mathsf {Ab}$). If $gf=0$, prove that $\ker g/ \text{im} f$ is isomorphic to the dual object $\text{coim} g /\text{coker} f$.



*

*How do I solve it?

*Why would one expect $\ker g/ \text{im} f$ to be "self dual", and what does it mean intuitively?

 A: Given arrows $A\xrightarrow{\ \ f\ \ }B\xrightarrow{\ \ g\ \ }C$ with $gf=0$, we can factor $f$ as $A\longrightarrow\operatorname{Im}f\longrightarrow\operatorname{Ker}g\longrightarrow B$. The object we have to deal with is the cokernel of the second  arrow in this sequence. Show that $\operatorname{coker}(\operatorname{Im}f\longrightarrow\operatorname{Ker}g)$ coincides with the coimage $\operatorname{coim}(\operatorname{Ker}g\longrightarrow B\longrightarrow\operatorname{Coker}f)$. Since these are both epimorphisms, it suffices to show that each of them factors through the other.
Then, since image and coimage coincide in abelian categories, we get 
$\begin{align*}\operatorname{Coker}(\operatorname{Im}f\longrightarrow\operatorname{Ker}g)&\cong\operatorname{Coim}(\operatorname{Ker}g\longrightarrow B\longrightarrow\operatorname{Coker}f)\\&\cong\operatorname{Im}(\operatorname{Ker}g\longrightarrow B\longrightarrow\operatorname{Coker}f)\\&\cong\operatorname{Ker}(\operatorname{Coker}f\longrightarrow\operatorname{Coim}g)\,,\end{align*}$
where the last isomorphism is formally dual to the first one. Thus the "self-duality" of the homology object (as one usually calls $\operatorname{Ker}g/\operatorname{Im}f$) follows directly from the fact that images are self-dual in abelian categories, which is nothing else than the usual homomorphism theorem from elementary module theory.
Edit: I was asked to carry out the argument mentioned in the first paragraph, so here we go: First note, that $\operatorname{coker}(\operatorname{Im}f\longrightarrow \operatorname{Ker}g)=\operatorname{coker}(A\longrightarrow\operatorname{Ker}g)$. To obtain our desired arrow $\operatorname{Coker}(\operatorname{Im}f\longrightarrow\operatorname{Ker}g)\longrightarrow\operatorname{Coim}(\operatorname{Ker}g\longrightarrow\operatorname{Coker}f)$, we have to check that $A\longrightarrow\operatorname{Ker}g\longrightarrow\operatorname{Coim}(\operatorname{Ker}g\longrightarrow\operatorname{Coker}f)$ vanishes. To do this, we are allowed to postcompose with a monomorphism, so it suffices to consider $$A\longrightarrow\operatorname{Ker}g\longrightarrow\operatorname{Coim}(\operatorname{Ker}g\longrightarrow\operatorname{Coker}f)\xrightarrow{\ \cong\ }\operatorname{Im}(\operatorname{Ker}g\longrightarrow\operatorname{Coker}f)\longrightarrow\operatorname{Coker}f\,.$$
By some commuting conditions, this is just the usual sequence $A\xrightarrow{\ \ f\ \ }B\longrightarrow\operatorname{Coker}f$. You should draw the corresponding diagram involving all these objects.
For the arrow in the reverse direction, let us use the universal property of the Coimage. We therefore have to factor $\operatorname{Ker}g\longrightarrow B\longrightarrow\operatorname{Coker}f$ through the epimorphism $\operatorname{Ker}g\longrightarrow\operatorname{Coker}(A\longrightarrow\operatorname{Ker}g)$. To do this, we have to check, that $A\longrightarrow\operatorname{Ker}g\longrightarrow B\longrightarrow\operatorname{Coker}f$ vanishes, which agian is obviously true.
